Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 7} = \dfrac{-3x - 2}{x - 7}$
Solution: Multiply both sides by $x - 7$ $ \dfrac{x^2}{x - 7} (x - 7) = \dfrac{-3x - 2}{x - 7} (x - 7)$ $ x^2 = -3x - 2$ Subtract $-3x - 2$ from both sides: $ x^2 - (-3x - 2) = -3x - 2 - (-3x - 2)$ $ x^2 + 3x + 2 = 0$ Factor the expression: $ (x + 2)(x + 1) = 0$ Therefore $x = -2$ or $x = -1$ The original expression is defined at $x = -2$ and $x = -1$, so there are no extraneous solutions.